Let $X}_{1},\cdots ,{X}_{n$ be a random sample from $f(x\mid \theta )=\theta {x}^{\theta -1}$ for $0<x<1$. Find a confidence interval for $\theta$ using as pivotal quantity a function of the maximum likelihood estimator for $\theta$.

dooporpplauttssg
2022-04-29
Answered

Let $X}_{1},\cdots ,{X}_{n$ be a random sample from $f(x\mid \theta )=\theta {x}^{\theta -1}$ for $0<x<1$. Find a confidence interval for $\theta$ using as pivotal quantity a function of the maximum likelihood estimator for $\theta$.

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Makayla Santiago

Answered 2022-04-30
Author has **19** answers

Step 1

Assuming you want a confidence interval for$\theta$ , you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that

$\hat{\theta}=-\frac{n}{\sum \mathrm{ln}\text{}{x}_{i}}.$

For the expected value we have the integral

$E\left[X\right]={\int}_{0}^{1}\theta {x}^{(\theta -1)}x\text{}\text{d}\theta =\frac{\theta}{1+\theta}$

And for the second moment

$\mu}_{2}={\int}_{0}^{1}\theta {x}^{\theta -1}{x}^{2}\text{}\text{d}\theta =\frac{\theta}{2+\theta$

So the variance is

$Var\left(X\right)=\frac{\theta}{2+\theta}-{\left(\frac{\theta}{1+\theta}\right)}^{2}=$

$=\frac{\theta}{{(\theta +1)}^{2}(\theta +2)}.$

Now then, we know that the following quotient has a chi-squared distribution, so in this case

$C=\frac{(n-1){S}^{2}}{{\sigma}^{2}}\sim {\chi}_{\{n-1\}}^{2}\text{}\text{and}\text{}C=\frac{(n-1){S}^{2}}{\frac{\theta}{{(\theta +1)}^{2}(\theta +2)}}$

where$S}^{2}=\frac{1}{n-1}\sum {({X}_{i}-\stackrel{\u2015}{X})}^{2$ . Finally we can work with the random variable C distributed as a chi squared. We want a confidence interval of $(1-\alpha )$ . Then

$P\left({\chi}_{\{n-1\}}^{2}\left(\frac{\alpha}{2}\right)<C<{\chi}_{\{n-1\}}^{2}(1-\frac{\alpha}{2})\right)=1-\alpha$

denoting that${\chi}_{\{n-1\}}^{2}\left(\beta \right)$ is the $\beta$ percentile of the $\chi}^{2$ which may be found in reference tables.

From here we work with the interval only substituting the percentiles with$\frac{1}{{D}_{t}}$ and $\frac{1}{{D}_{\pi}}$ , lower and superior.

$\frac{1}{{D}_{l}}<C<\frac{1}{{D}_{s}}\Rightarrow \frac{1}{{D}_{l}}<\frac{(n-1){S}^{2}}{\frac{\theta}{{(\theta +1)}^{2}(\theta +2)}}<\frac{1}{{D}_{s}}$

$D}_{l}>\frac{\frac{\theta}{{(\theta +1)}^{2}(\theta +2)}}{(n-1){S}^{2}}>{D}_{s$

${D}_{l}\left((n-1){S}^{2}\right)>\frac{\theta}{{(\theta +1)}^{2}(\theta +2)}>{D}_{s}\left((n-1){S}^{2}\right)$

$\left({D}_{l}\left((n-1){S}^{2}\right)\right)>\frac{\theta}{{(\theta +1)}^{2}(\theta +2)}>\left({D}_{s}\left((n-1){S}^{2}\right)\right)$

${(\hat{\theta}+1)}^{2}(\hat{\theta}+2)\left({D}_{l}\left((n-1){S}^{2}\right)\right)>\theta >{(\hat{\theta}+1)}^{2}(\hat{\theta}+2)\left({D}_{s}\left((n-1){S}^{2}\right)\right)$

where, taking the before definition and derivation,$\hat{\theta}=-\frac{n}{\sum \mathrm{ln}\text{}{x}_{i}}$ and $S}^{2}=\frac{1}{n-1}\sum {({X}_{i}-\stackrel{\u2015}{X})}^{2$ . When we sent the $\theta$ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include $\theta$ , then try it and tell us if it is simpler!

Assuming you want a confidence interval for

For the expected value we have the integral

And for the second moment

So the variance is

Now then, we know that the following quotient has a chi-squared distribution, so in this case

where

denoting that

From here we work with the interval only substituting the percentiles with

where, taking the before definition and derivation,

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Required sample size =

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Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.

Standard deviation = I have gotten max - min/4

Find the size of the sample required to estimage the mean age of textbooks currently used in college courses. Assume that you want 98% confidence that the sample mean is within 0.4 year of the population mean.

Required sample size =

I have no clue how to get the required sample size